So, in lieu of this post, I want to deepen the discussion of the Laurent series. We have the Laurent theorem:
If f(z) is analytic on $R_1\leq |z-z_o|\leq 2$, then $f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n$
Take for example the function $f(z)=\frac{1}{2-z}$, and develop it into series on the ring $0\leq|z|\leq 2$. First we see that the pole z=2 is within the region, so we develop this into a Power series (in the form of $\sum_{n=0}^{\infty}a_nz^n$:
\begin{equation} \frac{1}{2-z}=\frac{1}{2}\frac{1}{1-z/2}=\sum_{n=0}^\infty \bigg(\frac{1}{2}\bigg)^n(z/2)^n=\sum_{n=0}^\infty \frac{z^n}{2^{n+1}} \end{equation}
Then we have another case, where the same function is to be developed into a series in the region $0\leq|z|\leq1$: Here, clearly the pole $z=2$ is outside of this region, but since $|z-2|\leq 1$, it still satisfies the boundary of the region, so we still develop it into a Power series.
\begin{equation} \frac{1}{2-z}=\frac{1}{2}\frac{1}{1-z/2}=\sum_{n=0}^\infty \bigg(\frac{1}{2}\bigg)^n(z/2)^n=\sum_{n=0}^\infty \frac{z^n}{2^{n+1}} \end{equation}
Then we have the more in-obvious case, develop the function into a series on the annulus centered at $z_0=1$, that is $2\leq|z-1|\leq3$. This annulus has the pole of f(z) at its limit point, so we ideally still seek a Power series expansion, if I am not wrong. But since $|z-2|\leq 1 \notin 2\leq|z-1|\leq3 $, we have to develop this function into a Laurent series.
\begin{equation} \frac{1}{2-z}=\frac{1}{2}\frac{1}{1-z/2}= \frac{1}{2}\frac{1}{z}\bigg(-\frac{1}{1/2-1/z}\bigg) =\frac{1}{2}\frac{1}{z}\frac{1}{2}\bigg(-\frac{1}{1-2/z}\bigg) \end{equation}
Since we have the form of $a_n/z$ in the last term of the denominator, we can set up the Laurent series:
\begin{equation} -\frac{1}{2}\frac{1}{z}\frac{1}{2}\bigg(\frac{1}{1-2/z}\bigg)=\sum_{n=-\infty}^\infty(-1)^n\bigg(\frac{1}{2^2}\frac{1}{z}\frac{2}{z}\bigg)^n=\sum_{n=-\infty}^\infty(-1)^n\bigg(\frac{1}{2^{n+1}z^{n+1}}\bigg) \end{equation}
At this point, we need to respect the form of the boundary of the annulus and the theorem on top, and insert for z=|z-1| and correct the index.
\begin{equation} \sum_{n=-\infty}^\infty(-1)^{n-1}\bigg(\frac{1}{2^{n}(z-1)^{n}}\bigg) \end{equation}
Would this be correct, and if not, can you explain case by case , why?
Thanks!
That's not correct in what concerns the development into Lauren series in the annulus $2<|z-1|<3$. For such a $z$ we have\begin{align}\frac1{2-z}&=\frac1{1-(z-1)}\\&=-\frac1{z-1}\frac1{1-\frac1{z-1}}\\&=-\frac1{z-1}\sum_{n=0}^\infty\left(\frac1{z-1}\right)^n\\&=-\frac1{z-1}\sum_{n=0}^\infty(z-1)^{-n}\\&=-\frac1{z-1}\sum_{n=-\infty}^0(z-1)^n\\&=-\sum_{n=-\infty}^0(z-1)_{n-1}\\&=-\sum_{n=-\infty}^{-1}(z-1)^n.\end{align}